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2017 VCE Mathematical Methods 2 examination report
General comments
There were some excellent responses to the 2017 Mathematical Methods 2 examination and most
students were able to attempt the four questions in Section B. Question 1 was very well answered.
Some students found Questions 2f., 2g., 2h., 3g, 4h. and 4i. challenging.
Many students wrote their answers in the correct form. Exact answers needed to be given unless
otherwise specified. Approximate answers were usually required in the probability questions. Most
students answered everything that was required within a question. It is important for students to
re-read questions as often they require more than one piece of information in the response.
Advice to students
Technology should be updated well before the examination.
Technology is best put in mode radians. Occasionally the mode may need to be changed
according to the question, such as in Question 2d.
Check that answers make sense. By looking at the graph in Question 1bi., the gradient was
negative and likewise in Question 1d., the a value was positive.
Define functions on the technology at the start of each question in Section B. This saves time,
especially when dealing with probability questions that involve hybrid functions, such as
Question 3.
If the functions have been defined at the start of the question, it is acceptable to use the
function name, such as f(x), throughout the question rather than writing out the entire
expression. This avoids missing out on marks if brackets are not inserted when finding the
area between two curves, for example, in Question 1dii. and Question 4c. It also saves time
and avoids transcription errors.
Show a method for questions worth more than one mark. A small number of students did not
show their method in the probability question, Question 3.
Take time when sketching graphs, like in Question 3a. If it is a linear graph use a ruler. Check
that the points have been positioned correctly if a grid has been given. Sketch along an axis if
the function is defined for those points.
Use brackets with questions involving logarithms, such as Question 4b., 2log ( 2) 1y x .
Learn the correct wording to describe transformations. Not knowing this led to students making
errors in Question 4g.
Specific information
This report provides sample answers or an indication of what answers may have included. Unless
otherwise stated, these are not intended to be exemplary or complete responses.
The statistics in this report may be subject to rounding, resulting in a total more or less than 100 per cent.
2017 VCE Mathematical Methods 2 examination report
© VCAA Page 2
Section A The table below indicates the percentage of students who chose each option. The correct
answer is indicated by the shading.
Question % A % B % C % D % E % No
answer Comments
1 4 1 92 2 1 0
2 1 5 12 80 1 0
3 2 2 83 12 1 0
4 3 10 6 7 75 0
5 47 20 9 16 7 1
95% confidence interval is (0.039, 0.121).
The sample proportion is in the middle of the confidence interval.
0.121 0.0390.039 0.080
2
6 5 3 88 3 1 0
7 19 32 12 29 7 1
2( 1) 4 5p x x p
2( 1) 4 5 0p x x p
The discriminant is negative for no real solutions.
16 4( 1)( 5) 0p p
24 24 4 0p p
Divide by 4 and change the inequality.
2 6 1 0p p
8 64 14 6 11 5 0
9 78 4 6 8 4 0
10 9 23 6 47 14 0
2 0
10
3
x xT
y y
2 , 2
xx x x
1, 3
3y y y y
3sin 24
y x
3 3sin 22 4
xy
sin2
y x
cosy x
11 10 5 10 72 3 0
2017 VCE Mathematical Methods 2 examination report
© VCAA Page 3
Question % A % B % C % D % E % No
answer Comments
12 11 18 45 15 11 1
3sin(2 )
2x
22 , ...
3 3x
, ...6 3
x
5 2, , ,
6 3 6 3x
5 2sum:
6 3 6 3
13 12 14 17 10 46 1
1( )
1h x
x
2 2( )h x h x
2
2 2
1 1 1
1 2 1 1x x x x
14 14 10 9 62 4 1
15 14 58 7 9 11 0
16 12 15 41 17 13 2
Bi 5,X p
0 55 1 2
Pr 0 (1 ) , 0 243 3
X p p p
Pr 3 Pr 4 0.4609X X , correct to four
decimal places
2017 VCE Mathematical Methods 2 examination report
© VCAA Page 4
Question % A % B % C % D % E % No
answer Comments
17 3 37 21 21 17 0
Area = ( ) ( ) ( )
b c d
a b c
f x dx f x dx f x dx
2 ( ) ( )
b c
a b
f x dx f x dx
2 ( ) 2 ( )
b b c
a b
f x dx f x dx
, as
0, since ( ) ( )b c f x f x
18 10 16 25 38 9 2
(1 )np np p
2 2 (1 )n p np p
1 0, 0np np p np
11 ,
1np p p
n
10.01, 99
1n
n
19 6 17 9 59 8 1
20 8 47 18 18 9 1
cos( ) 3sin( )x x
1tan( ) ,
63x x
3,
6 2B
triangle
3 3
4 2 8A
6 2
shaded
0
6
3sin( ) cos( ) 3 1A x dx x dx
shaded triangle:A A
3
3 1:8
2017 VCE Mathematical Methods 2 examination report
© VCAA Page 5
Section B
Question 1a.
Marks 0 1 2 Average
% 7 19 74 1.7
3: , ( ) 5f R R f x x x , solve ( ) 0f x for x, 15
3x , turning points
15 10 15 15 10 15, , ,
3 9 3 9
This question was answered well. Exact answers were required to obtain full marks. Some
students gave their answers as (–1.29, 4.3) and (1.29, –4.3). Others gave only the x values. Some
mixed up the signs.
Question 1bi.
Marks 0 1 2 Average
% 16 9 75 1.6
gradient (1) ( 1)
41 1
f f
, 4 4( 1)y x , 4y x
This question was answered well. A common incorrect answer was 4y x . By inspection of the
graph, the gradient was negative. Some students made arithmetic errors when calculating the
gradient and/or y-intercept.
Question 1bii.
Marks 0 1 Average
% 22 78 0.8
22
2 1 2 1( ) ( ) ( )d x x f x f x 22(1 1) (1) ( 1)f f 68 2 17
The distance formula was used well. Some students applied this by hand and made arithmetic
errors. Others had an incorrect distance formula using a multiplication operation between the two
brackets instead of a plus. A common incorrect answer was 64 .
Question 1ci.
Marks 0 1 2 Average
% 17 20 64 1.5
22
2 1 2 1( ) ( ) ( )d x x g x g x 22 2(1 ( 1)) (1) ( 1) 2 2 2 g g k k
As in Question 1bii. some students did their solutions by hand and made arithmetic errors,
especially sign errors. This would have been time consuming. kk 222)22(2 22 was
sometimes given. Some incorrect answers contained . When defining 3( )g x x kx on the
technology, a multiplication sign must be inserted between k and x.
2017 VCE Mathematical Methods 2 examination report
© VCAA Page 6
Question 1cii.
Marks 0 1 Average
% 37 63 0.7
Solve 22 2 2 1k k k for k, 7
1 or 3
k k
Students who answered Question 1ci. correctly were generally able to answer this question. Some
students gave only one value for k.
Question 1di.
Marks 0 1 Average
% 38 62 0.6
Solve ( )g x x for x, 1a k , as a > 0
A common incorrect answer was 1a k . By inspection of the graph, the answer was positive.
Some students found k in terms of a, instead of a in terms of k.
Question 1dii.
Marks 0 1 2 Average
% 40 18 42 1
1 2
0
( 1)( )
4
kk
x g x dx
1
3
0
k
x x kx dx
was a common error, leaving out the brackets in 1
3
0
( )k
x x kx dx
. To avoid
these errors it would have been better to use the expression 1
0
( )k
x g x dx
. Some students
overcomplicated the question by breaking up the areas into different sections. The easiest approach was to use ‘upper function subtract lower function’. There was evidence that students
substituted 1 k instead of 1k and this resulted in the answer of 4
)1)(3( kk.
Question 2a.
Marks 0 1 Average
% 11 89 0.9
( ) 65 55cos15
th t
, range 55 65,55 65 10,120 , minimum height is 10 m and maximum
height is 120 m
This question was well answered.
2017 VCE Mathematical Methods 2 examination report
© VCAA Page 7
Question 2b.
Marks 0 1 Average
% 16 84 0.9
Period = 2
30
15
. He was in the capsule for 30 minutes.
A common incorrect answer was 15 minutes.
Question 2c.
Marks 0 1 2 Average
% 29 42 29 1
11( ) sin
3 15
th t
, solve
11( )
3h t
for t, t = 7.5 minutes
Most students were able to find the derivative. There were occasions when 12
12
xx
yy
was attempted
(average rate of change). Some students had their technology in degree instead of radian mode,
giving
211 sin15
( )540
t
h t
. Many could not find the maximum rate of change. A common incorrect
answer was 15 minutes. Many found the value of t for the maximum value of h. Others gave a
general solution or two t values.
Question 2d.
Marks 0 1 Average
% 64 36 0.4
65tan( ) , 7.41
500 , correct to two decimal places
Many students knew to get
500
65tan 1
but they did not specify ‘degree’ for their technology. A
common incorrect answer was 1 55tan 6.28
500
. Some used 65
tan500
instead of
1 65tan
500
. Others used 1 65sin
500
.
Students should be familiar with the relevant functionality for the context and select it appropriately. Question 2e.
Marks 0 1 Average
% 10 91 0.9
23025
dy x
dx x
2017 VCE Mathematical Methods 2 examination report
© VCAA Page 8
This question was answered well. Some students wrote 23025
dy x
dx x
. There was no need to
rationalise the denominator. Question 2f.
Marks 0 1 2 3 Average
% 67 21 4 8 0.5
2
2
2
3025 65 or
5003025P B
u um
uu
, solve
2
2
3025 65 =
5003025
u u
uu
for u
12.9975... 13.00u , 118.4421... 118.44v , correct to two decimal places
Many students were able to find the gradient of the line segment in terms of u, using their answer
from Question 2e. Others used h(t) or 23025y x instead of
23025 65y x . Many
students were unable to find the second gradient expression where they were required to use rise
over run for the line segment P2B.
Question 2g.
Marks 0 1 Average
% 93 7 0.1
1
2
12.9975...tan 13.67
3025 12.9975...
, correct to two decimal places
Some students used radians instead of degrees. Question 2h.
Marks 0 1 2 Average
% 94 5 2 0.1
Angle difference = 90 (13.669... 7.406...) 83.737... ,83.737...
30 6.978...360
7 minutes, to the
nearest minute
This question was not answered well. Some students wrote 7 minutes without showing any
working. As indicated in the instructions on the examination, for questions worth more than one
mark, appropriate working must be shown.
2017 VCE Mathematical Methods 2 examination report
© VCAA Page 9
Question 3a.
Marks 0 1 2 3 Average
% 15 12 44 29 1.9
Many students did not draw their graphs along the t-axis, ignoring f(t) = 0. Some had an open circle
at (45, 0.04). Others had an open circle over a closed circle at (45, 0.04). Many students did not
use rulers to draw the line segments. Some graphs looked like parabolas.
Question 3b.
Marks 0 1 2 Average
% 24 6 70 1.5
55
25
4( )
5f t dt
This question was answered well. Some students had the incorrect terminals. 44 instead of 45 was
occasionally given, for example, 44 55
25 44
( ) ( ) f t dt f t dt . Others used 20 as the lower limit instead
of 25.
Question 3c.
Marks 0 1 2 Average
% 21 30 49 1.3
Pr 25Pr 25 | 55
Pr 55
TT T
T
25
20
55
20
( )
( )
f t dt
f t dt
1
41
Many students were able to use the conditional probability formula. A common incorrect answer
was 1
40.
2017 VCE Mathematical Methods 2 examination report
© VCAA Page 10
Question 3d.
Marks 0 1 2 Average
% 59 11 30 0.7
70 45
20
( ) 0.7 or ( ) 0.3 or ( ) 0.2, 39.3649
a
a a
f t dt f t dt f t dt a , correct to four decimal places
A number of correct approaches were used. 20
( ) 0.7
a
f t dt , a = 50.6351 was a common incorrect
answer. 75
170 0.7
625a
t dt was occasionally given. Some students attempted to use the inverse
normal as a method.
Question 3ei.
Marks 0 1 2 Average
% 26 16 58 1.3
8Bi 7,
25X
, Pr 3 0.1534X , correct to four decimal places
Many students recognised that the distribution was binomial and gave the correct n and p values.
Some used Pr 3X .
Question 3eii.
Marks 0 1 2 Average
% 31 11 59 1.3
Pr 2Pr 2 | 1
Pr 1
XX X
X
0.7113...0.7626
0.93277... , correct to four decimal places
Many students were able to set up the conditional probability. Some wrote
Pr 2Pr 2 | 1
Pr 1
XX X
X
. Others rounded incorrectly, giving 0.7625 as the answer.
Question 3f.
Marks 0 1 2 Average
% 60 8 32 0.7
2 5 3 47 7( ) (1 ) (1 )
2 3
q p p p p p2 47 ( 1) (2 3)p p p
Of those who attempted this question, some students did not realise that the binomial distribution
was required.
Question 3gi.
Marks 0 1 2 Average
% 64 13 23 0.6
Solve ( ) 0q p for p, 0.3539, 0.5665p q , correct to four decimal places
2017 VCE Mathematical Methods 2 examination report
© VCAA Page 11
Some students knew to solve ( ) 0q p if they had an equation in Question 3f. Others found only
p. Some gave exact values for their answers.
Question 3gii.
Marks 0 1 2 Average
% 91 3 7 0.2
70
( ) 0.35388...
d
f t dt 48.9676....d 49 minutes, to the nearest minute
Some students used q instead of p in their equation, solving
70
( ) 0.56646...
d
f t dt for d. Others
solved
20
( ) 0.35388...
d
f t dt , obtaining d = 41 minutes.
Question 4a.
Marks 0 1 2 Average
% 24 28 48 1.3
1( ) 2 2xf x , the graph of 2xy has been translated one unit to the left and two units down to
get the graph of f, 1, 2c d or x x c
Ty y d
, ,x x c x x c , ,y y d y y d ,
2 , 2x c x cy d y d , 1, 2c d
This question was answered reasonably well. Some students made sign errors.
Question 4b.
Marks 0 1 2 Average
% 12 25 64 1.5
Let 12 2xy , inverse swap x and y, 12 2yx , 12 2yx , 2log ( 2) 1y x
1 12: 2, , ( ) log ( 2) 1f R f x x
There are other acceptable expressions for the inverse. Some students found the rule but did not
give the domain. Some students did not use brackets, leaving their answer as 2log 2 1y x .
2017 VCE Mathematical Methods 2 examination report
© VCAA Page 12
Question 4c.
Marks 0 1 2 3 Average
% 22 10 20 49 2
Solve1( ) ( ), 1 or 0f x f x x x ,
0 0
1
1 1
22 ( ) ( ) ( ) 3
log (2)e
x f x dx f x f x dx
0
1
1
( ) ( )f x f x dx
and 0
1 1
1
( ) 2 2xf x dx
were common incorrect expressions.
0
1 1
1
( ) 2 2xf x dx
should be written as 0
1 1
1
( ) (2 2)xf x dx
. To avoid this type of error it
is better to use the expression 0
1
1
( ) ( )f x f x dx
. An exact answer was required. 0.1196 was
often given.
Question 4d.
Marks 0 1 2 Average
% 26 13 62 1.4
(0) 2log (2)ef , 1 1(0)
2log (2)e
f
This question was answered reasonably well.
Question 4e.
Marks 0 1 Average
% 35 65 0.7
( ) 2 2kxkg x e , solve ( ) ( )kg x f x for k, log (2)ek
This question was answered reasonably well.
Question 4f.
Marks 0 1 Average
% 42 58 0.6
1 1 2( ) log
2k e
xg x
k
Some students used their answer to Question 4e. 1 1 2( ) log
log (2) 2ek
e
xg x
was a common
incorrect response.
Question 4gi.
Marks 0 1 Average
% 69 31 0.3
1( ) 2 2xg x e , ( ) 2 2kxkg x e , dilation of a factor of
1
k from the y-axis
© VCAA Page 13
Many students were unable to describe the transformation correctly, for example, ‘dilation of a
factor 1
k in the y-axis’. Others put their answer in terms of log (2)e instead of k. Some gave two
transformations.
Question 4gii.
Marks 0 1 Average
% 70 30 0.3
11
2( ) log
2e
xg x
, 1 1 2( ) log
2k e
xg x
k
, dilation of a factor of
1
k from the x-axis
Students who answered Question 4gi. correctly tended to answer this question well.
Question 4h.
Marks 0 1 2 Average
% 77 21 2 0.3
( ) 2 2kxkg x e , 1 1 2
( ) log2
k e
xg x
k
, 1 2: 2 , :
2
xL y kx L y
k ,
1
2 tan 60 3 or 2 tan 303
k k 3 1
or 2 2 3
k k
This question was not answered well. Some students found one answer only. Others gave
approximate answers.
Question 4ii.
Marks 0 1 2 Average
% 95 2 3 0.1
Solve1
22
kk
for k, 1
, 02
k k , 1
2p
Many students tried to solve 1( ) ( )k kg x g x and then attempted to find the discriminant.
Question 4iii.
Marks 0 1 Average
% 98 2 0
As 1
2k the graphs of
1 and k kg g will intersect in the third quadrant,
0
2
lim ( ) 4k x f x dx
, 4b ,
as 1
kg has a vertical asymptote with equation 2x and kg has a horizontal asymptote with
equation 2y , the area will approach 4 as k increases.
This question was not answered well.
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